TENSION FORCE QUESTIONS IN ROPE III

Physics lesson, tension force in rope. Frictional and tensional forces on an inclined plane. Questions and its solutions.

Question -1

On the inclined plane seen in the figure the K and L objects are balanced by the F force

The mass of the K object is 10 kg, and the mass of the L object is 15 kg. The coefficient of friction is 0,2. The system is at rest.

A) Find the F force that keeps the system at rest.

B) Find the acceleration of the system when the F force is 100 N.

C) If in case of hanging an P object of 40 kg the end of the rope instead of the F force, what will be the tension of the ropes T1 and T2 and the acceleration of the system?

(g = 10 m/s2, sin37° = 0.6, cos 53° = 0.8)

Solution – 1

The forces that effect to the objects is having been shown above.

The F force tries to pull upward the K and L objects. The gravity force tries to pull its downward.

The FK and FL forces are parallel components of the gravity to the inclined plane surface.

The force that enforce slip the object downward on the inclined plane is gravity force.

FK and FL can be found from the trigonometric ratios.

Also there is a friction force in the system. The friction force is can be found as below.

Fs = N•k

N: Normal of the surface

k: Friction coefficient

Normal of the surface be found as below;

N = m•g

If angle between earth and the surface that the object is on is not zero, normal of the surface be found as below;

N = m•g•cosϑ

The direction of friction force is always opposite the move direction of the system. Let’s first find the friction force, then the FK and FL forces.

The normal of K object is equal to the "WKy"

WKy = 0.6 • WK

WK = mK•g

WK = 10 • 10

WK = 100 N

WKy = 0.6 • 100

WKy = 60 N

The friction force that affect to the K object.

FsK = NK•0.2

FsK = 60 • 0.2

FsK = 12 N

The normal of L object is equal to "WLy";

WLy = WL • cos53

WL = 15 • 10

WL = 150 N

WLy = 150 • 0.6

WLy = 90 N

The friction force that affect to the L object;

FsL = NL • 0.2

FsL = 90 • 0.2

FsL = 18 N

The total friction force which is the affect on the K and L object.

Fs = FsK + FsL

Fs = 12 + 18

Fs = 30 N

The FK force that pulls the K object down.

FK = WKx

WKx = WK • sin53°

WKx = 100 • 0.8

WKx = 80 N

FK = 80 N

The FL force that pulls the L object down,

FL = WLx

WLx = WL • sin53°

WL = mL•g

WL = 15 • 10 = 150 N

WLx = 150 • 0.8

WLx = 120 N

The sum of the forces that enforce the system to move in "1" direction.

FK + FL = 80 + 120 = 200 N

In the absence of the F force, the system is will move in the "1" direction.

When the system is move in direction the "1" the friction force is will in direction the "2". So, the sum of the F and friction forces are prevent the move of system in direction the "1".

F + Fs = FK + FL

F + Fs = 200 N

F + 30 = 200 N

F = 170 N

The force applied to keep the system remain at rest must be at least 170 N.

B)

The system will start to move in direction the “1” when the F force is 100 N. Because the F force is is less than FK and FL forces.

Fnet = FK + FL – Fs – F

Fnet = 80 + 120 – 30 – 100

Fnet = m.a

70 = (10 + 15)•a

70 = 25•a

a = 2.8 m/s2

The acceleration of the system is 2.8 m/s2.

C)

If is hang a 40 kg object to the end of the rope the force that the object will apply to the rope is equal to the weight of the object.

WP = 40 •10

Wp= 400 N

The system will start to move in direction of the "2" because of the force of WP is greater than the sum of the forces FK, FL and FS.

Fnet = m.a

FP – 80 – 120 – 30 = (10 + 15 + 40).a

170 = 65•a

a = 2.6 m/s2

Now, let we find the tension forces of T1 and T2 ropes.

For the T2 rope,

The forces that affected the object are FsK (friction force), FK and T2 forces.
The T2 force pulls the object K upward. The FK force is pulls the object K downward. The friction force (FsK) is affect in direction "2" because of the direction of movement of the system is "1"

We will do analysis on the K object for found the T2 rope tension.

Fnet = m•a

Fnet = 10 • 2.6

Fnet = 26 N

Fnet = T2 – FsK – FK

Fnet = T2 – 12 – 80

26 = T2 – 92

T2 = 118 N

For T1 rope,

The forces that pulls the L object in the direction "2" are FK and FL forces. The force that pulls the L object in the direction of "1" is the T1 forces.

When we want to find tension of the T1 rope we analyze on the L object.

Fnet = m•a

Fnet = 15 • 2.6

Fnet = 39 N

Fnet = T1 – T2 – FsL – FL

Fnet = T1 – 118 – 18

39 = T1 – 136

T1 = 175 N

Question – 2

The mass of the P object which in the figure is 9 kg and it is balanced with T rope.

A) Find the tension force in rope.

B) Find the force that the object apply the Wall.

Solution - 2:

The forces that are effective on the P object is shown at below.

The m.g force is pulls the P object down. The Tx force is the force applied by P to the Wall. The Ty force is force that pulls the object upward.

Ty

= cos53°

T

Also the m.g vector is equal to the vertical component of T vector.

Ty = m•g

m•g = 9•10 = 90 N

90

= 0.6

T

T =

90

0.6

T = 150 N

B)

The force that the object apply the wall is equal to the Tx component.